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Cambridge IGCSE™ Chemistry
Cambridge IGCSETM Chemistry Workbook answers Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers which are contained in this publication.
1 States of matter Core 1
In ice, I (the water molecule) am bonded to other water molecules and vibrating in a fixed position. As the temperature rises, there is enough energy for some of the bonds between us to weaken and begin to break due to the increase in vibrations. This allows me to move further from my neighbouring molecules and I am able to move past them. As the temperature continues to rise, more of the bonds between us begin to break and I move away from my neighbouring molecules and move much more quickly. At 100 °C I have so little attraction to my neighbouring molecules that I am able to break away and move a long way away from them.
2
a
i
When a substance changes from a solid to a liquid
ii
When substances are changed into different substances by chemical reaction
iii When a gas changes into a liquid iv When a liquid changes into a gas v
When a liquid changes into a solid
vi When heating of a liquid causes it to vaporize b 3
i
Condensation
ii
Melting
a
Diffusion cannot occur in solids because the particles are bonded to one another more strongly than those in liquids and gases, which do diffuse. This prevents them from being able to move relative to one another.
b
Gases can be compressed more than liquids because the particles in the gas are much more spread out than those in a liquid. This means that the gas particles can be squeezed together more than those in a liquid, which are much closer to one another.
c
Liquids can flow because their particles have much weaker bonds between them than those in a solid. This means that the particles in a liquid can move over one another.
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Cambridge IGCSE Chemistry Workbook answers
Supplement 4
5
a
[Correct scale, axis labels and units, plotting of points, and line]
b
5 °C
c
82 °C
a
Ammonia
b
Point B. Both HCl and NH3 particles move randomly along the tube, but the ammonia particles are lighter and so will move faster and further along the tube than the HCl particles.
c
The particles of hydrogen chloride and ammonia move along the tube by randomly colliding with other particles and the sides of the tube; the particles of ammonia move quicker as they are lighter.
Exam-style questions Core 1
a
i
Solid – any two of: cannot be compressed, has a fixed shape, expands on heating [2] Liquid – any two of: will flow and take up the shape of the container, can only be compressed a small amount, its volume increases when heated [2] Gas – any two of: is easily compressed, flows easily, will fill the whole volume of its container [2]
ii
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Cambridge IGCSE Chemistry Workbook answers
[1 for each] b
i
oxygen [1]
ii
phosphorus and iron [1]
iii bromine [1] 2
a
Gas particles from the coffee are moving randomly from the coffee shop, colliding with other particles in the air, until they reach you. [1] They are diffusing. [1]
b
When the temperature rises, the steel tracks will expand. [1] The gaps allow the tracks to expand without buckling the railway lines. [1]
c
The particles of tea move randomly and collide with water molecules in the water. [1] They diffuse from the tea bag until eventually all of the water in the cup changes colour as the tea dissolves. [1]
d
Water vapour inside the house, caused by central heating, hot water and people breathing [1], condenses onto the glass of the windows, which is colder. [1]
e
At the bottom of the ocean the bubble has a large amount of pressure all around it due to the depth of water, and has an equal interior pressure caused by methane molecules. [1] As the bubble rises, the external pressure decreases and the relative greater interior pressure of the gas in the bubble causes the volume of the methane bubble to increase. [1]
Supplement 3
a
i
The external pressure decreases. [1]
ii
The volume of the balloon will increase. [1]
iii As the external pressure decreases the particles of helium gas on the inside are still colliding with the inside surface [1] of the balloon, causing the balloon to increase in volume. [1]. b
i
The external temperature decreases. [1]
ii
As the external temperature decreases this will have an effect of cooling down the gas in the balloon and the particles will lose kinetic energy [1] and will cause them to collide with the inside surface of the balloon less violently [1] and this will lead to the balloon’s decreasing in volume. [1]
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Cambridge IGCSE Chemistry Workbook answers
2 Atoms, elements and compounds Core 1
a
The elements in a compound, e.g. iron sulfide, cannot be separated by physical means, whilst a mixture of iron filings and sulfur powder can be separated using a magnet.
b
i
carbon monoxide, sulfuric acid, methane, sodium hydroxide, limestone
ii
stainless steel, lemonade, cement, beer, brass
c
Accept any sensible answer with a reasonable explanation, for example: i
water, water is a compound
ii
chromium, chromium is an element
iii F2, F2 is an element 2
d
A group of atoms chemically joined/bonded together.
a
i
Use eye protection e.g. safety spectacles to protect eyes.
ii
A new substance is being produced.
iii zinc sulfide iv zinc + sulfur → zinc sulfide Zn(s) + S(s) → ZnS(s) [Correct reactants, correct products]
b
v
Zinc and sulfur can be separated by physical means (e.g. using an organic solvent to dissolve the sulfur) whereas the compound zinc sulfide needs chemical means to separate the elements.
i
copper, oxygen, carbon
ii
copper sulfide, copper oxide, sulfur dioxide, carbon dioxide
iii copper sulfide + oxygen → copper oxide + sulfur dioxide 2CuS(s) + 3O2(g) → 2CuO(s) + 2SO2(g) [Correct reactants, correct products, correct balancing] copper oxide + carbon → copper + carbon dioxide 2CuO(s) + C(s) → 2Cu(s) + CO2(g) [Correct reactants, correct products, correct balancing] 3
a
Formula of substance
LiNO3
Elements present
Total number of atoms
Symbol
Name
Number of atoms
Li
Lithium
1
N
Nitrogen
1
O
Oxygen
3
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4
Cambridge IGCSE Chemistry Workbook answers
CaCO3
Mg3N2
Ag2CrO4
b
Ca
Calcium
1
C
Carbon
1
O
Oxygen
3
Mg
Magnesium
3
N
Nitrogen
2
Ag
Silver
2
Cr
Chromium
1
O
Oxygen
4
i
2Pb(s) + O2(g) → 2PbO(s) [Correct products, correct reactants]
ii
2H2(g) + O2(g) → 2H2O(l) [Correct products, correct reactants]
5
5
7
iii C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l) [Correct products, correct reactants] 4
a Element
Proton number
Lithium
3
Boron Aluminium
a
6
5
20
i
23 11Na: 2, 8, 1
ii
1632 32 16S: 2, 8, 6
iii
39 19K: 2, 8, 8, 1
Nucleon number
27 7
Calcium
5
Number of electrons 3
13
Nitrogen
b
Number of neutrons
7 20
12 × mass of 1 atom of carbon-12 Ar = average mass of the isotopes of an element compared to 1/12th of the mass of an atom of 12C.
b
Average mass = Ar =
c
(1.10 × 13) + (98.90 × 12) = 12.01 100
12.01 = 12.01 1
These are atoms of the same element, C. They have different nucleon numbers. C-12 has 6 neutrons and C-14 has 8 neutrons. Both atoms have 6 protons, hence they are isotopes of carbon.
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Cambridge IGCSE Chemistry Workbook answers
d
These are atoms of the same element with the same chemical properties because they have the same number of electrons and therefore the same electronic configuration.
Supplement 6
a
An oxidising agent is a substance that oxidises another substance and is itself reduced. A reducing agent is a substance that reduces another substance and is itself oxidised.
b
i
ZnO or zinc oxide
ii
C or carbon
Exam-style questions Core 1
a
Element
Metal or nonmetal?
W
Shiny?
Conductor of electricity?
Melting point
Yes
X Y
Metal
Z
No
b
high density
c
i
iron or oxygen
ii
iron oxide
iii 4Fe(s) + 3O2(g) → 2Fe2O3 (s) [Correct reactants, correct products, correct balancing]
Supplement 2
a
i
C(s) + O2(g) → CO2(g) [Correct reactants, correct products]
ii
CO2(g) + C(s) → 2CO(g) [Correct reactants, correct products, correct balancing]
b
carbon
c
carbon dioxide [1]
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Cambridge IGCSE Chemistry Workbook answers
d
Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g) [Correctly balanced reactants, correctly balanced products]
e
reducing agent CO, oxidising agent Fe2O3
3 Bonding and structure Core 1
2
a
false
b
true
c
false
d
true
e
true
f
true
a
A diamond, B graphite i
covalent
b
ii
giant covalent
c
i
4
ii
3
d
graphite – lubricant or electrode, diamond – cutting tools
e Substance
Electrical conductivity Melting point
Hardness
A
Poor
High
High
B
Good
High
Low
3
a
Ionic bonds are usually found in compounds that contain metals combined with nonmetals. When this type of bond is formed, electrons are transferred from the metal atoms to the non-metal atoms during the chemical reaction. In doing this positive metal ions and negative non-metal ions are formed. The positive ions are known as cations. The negative ions are known as anions. When oppositely charged ions form a crystal structure they have a strong attractive force between the them called an electrostatic force of attraction. This is known as the ionic bond.
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Cambridge IGCSE Chemistry Workbook answers
b
[Each metal and non-metal ion shown correctly, including charge] i
ii
4
LiCl
CaS
[In each, correct numbers of electrons, and correct sharing in overlap areas] a
HF
b
NCl3
Supplement 5
a
i
CuF2
ii
Na2CO3
iii Ag3PO4
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Cambridge IGCSE Chemistry Workbook answers
iv (NH4)2SO4 v
Mg3(PO4)2
vi Al(OH)3 vii FeBr3 6
Metals are good conductors of electricity and heat, because the free electrons from the outer energy levels of metal atoms carry a negative charge or heat energy through the metal. The free electrons are often described as delocalised. The free electrons allow metal ions to slide over each other, so metals are malleable and ductile. They have high melting and boiling points due to the strong attractive forces within the structure of the metal.
Exam-style questions Core 1
a
covalent (single bonds) [1]
b
electrons [1]
c
In the molecule, the carbon atom has eight electrons in its outer energy level. [1] It achieves this by sharing electrons with four hydrogen atoms. [1]
d
i
helium [1]
ii
neon [1]
iii The only shell in the hydrogen atom is the first, which is full when it contains only two electrons. [1]
Supplement 2
a
Na: 2,8,1 [1], Na+: 2,8 [1]
b
Cl: 2,8,7 [1], Cl–: 2,8,8 [1]
c
ionic [1]
d
i
Each Na+ ion is surrounded by six Cl– ions [1] and vice versa. [1]
ii
There are strong electrostatic forces between the oppositely charged ions. [1] A lot of energy is therefore needed to separate the ions and melt the substance. [1]
iii In the solid state the ions are not free to move [1] but in the molten state the ions are free to move to the oppositely charged electrodes. [1]
4 Stoichiometry – chemical calculations Core 1
a
The molecular formula of a compound is the number and type of different atoms in one molecule.
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Cambridge IGCSE Chemistry Workbook answers
b
A CH4;
B C2H6O;
2
g/dm3
mol/dm3
3
a
i
46
ii
88
C C2H4O2
iii 74 iv 42 b
i
106
ii
74
iii 132
Supplement 4
a
i
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g) [Correct reactants, correct products, correct balancing]
ii
moles of Fe used = 5.6/56 = 0.1 mole
iii moles of FeCl2 = 0.1 mole; mass of FeCl2 = 0.1 × (56 + 35.5 + 35.5) = 12.7 g iv percentage yield = (9.17/12.7) × 100 = 72.2% b
1 mole of Fe2O3 = (56 × 2) + (16 × 3) = 160 g number of moles of Fe2O3 = 100 × 106/160 = 625 000 moles moles of Fe formed, if 100% yield = 2 × 625 000 = 1 250 000 moles mass of Fe formed if 100% yield = 1 250 000 × 56 = 70 000 000 g (70 tonnes) percentage yield = (7/70) × 100 = 10%
5
6
7
8
a
i
2 moles
ii
0.01 mole
b
0.5 mole
c
3.01 × 1023 sodium ions and 3.01 × 1023 chloride ions = 6.02 × 1023 ions
a
i
200g
ii
4.5g
b
100
c
56 g/mol
a
The empirical formula of a compound is the simplest whole number ratio of the different elements in a compound.
b
2.16/12 = 0.18 mole carbon, 0.36/1 = 0.36 mole hydrogen, 1.44/16 = 0.09 mole oxygen, so two C and four H for every one O, C2H4O
c
Every 100 g has 27.4/23 = 1.19 moles sodium, 1.2/1 = 1.2 moles hydrogen, 14.3/12 = 1.19 moles carbon, 57.1/16 = 3.56 moles oxygen; approximately a ratio of 1 : 1 : 1 : 3, so NaHCO3
a
i
0.125 mole
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Cambridge IGCSE Chemistry Workbook answers
b 9
ii
0.05 mole
i
1 mol/dm3
ii
1 mol/dm3
a
2HCl + Na2CO3 → 2NaCl + H2O + CO2 [Correct reactants, correct products, correct balancing]
b
[five (or more) from the following]
c
•
25 cm3 of the sodium carbonate solution is placed in a conical flask using a pipette and safety filler.
•
3 or 4 drops of thymolphthalein (or methyl orange) indicator are added to the sodium carbonate solution.
•
A burette is filled with the hydrochloric acid solution, ensuring that some runs through the valve, and the initial burette reading is taken.
•
The acid is added from the burette into the flask, with swirling, until the colour of the indicator just changes.
•
The final burette reading is taken and the volume of acid needed to neutralise the sodium carbonate is found.
•
The process is repeated to obtain three concordant results (within 0.10 cm3 of one another).
moles of Na2CO3 used = 0.1 × 25/1000 = 2.5 × 10–3 mole moles of HCl = 2 × 2.5 × 10–3 = 5 × 10–3 mole concentration of HCl = 5 × 10–3 × 1000/18.95 = 0.26 mol/dm3
Exam-style questions Supplement 1
a
i
moles of CuO used = 8/(64 + 16) = 0.1 mole [1] moles of H2 gas needed = 0.1 mole [1] volume of H2 gas = 0.1 × 24 = 2.4 dm3 (2400 cm3) [1]
ii
moles of Cu obtained = 0.1 mole [1] mass of Cu = 0.1 × 64 = 6.4 g [1]
iii percentage yield = 5.8 × 100/6.4 [1] = 90.6% [1] b
i
moles of propane = 10/24 = 0.417 mole [1] moles of O2 needed = 5 × 0.417 = 2.085 mole [1] volume of O2 needed = 2.085 × 24 = 50 dm3 [1]
ii
moles of O2 = 10/24 = 0.417 volume of CO2 = 0.417/5 × 3 × 24 = 6 dm3 [1] volume of H2O = 0.417/5 × 4 × 24 = 8 dm3 [1] total volume = 6 + 8 = 14 dm3 [1]
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Cambridge IGCSE Chemistry Workbook answers
2
a Rough
1
2
3
21.75
22.25
22.35
22.30
Final burette reading/cm3 Initial burette reading/cm3 Volume of sulfuric acid used/cm3
[1] b
average volume = (22.25 + 22.35 + 22.30)/3 [1] = 22.30 cm3 [1]
c
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) [1 for reactants, 1 for products, 1 for correct balancing]
d
moles of NaOH = 0.25 × 25/1000 [1] = 6.25 × 10–3 mole [1]
e
moles of H2SO4 neutralised = 0.5 × 6.25 × 10–3 = 3.125 × 10–3 mole [1]
f
concentration of H2SO4 = 3.125 × 10–3 × 1000/22.30 [1] = 0.14 mol/dm3 [1]
5 Electrochemistry Core 1
a
substances T, U, W and Z
b
substances X and Y
c
substance U
d
substance X
e
substance V
f
any dilute acid or dilute alkali
g
copper(II) oxide
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Cambridge IGCSE Chemistry Workbook answers
2 Material electrodes
Substance
of Substance formed at Substance formed at the cathode the anode Lead
Chlorine
Calcium
Bromine
Dilute sulfuric acid Carbon Molten sodium chloride
Carbon
3
Electroplating is the process that uses electrolysis to plate (or coat) one metal with another. Often the purpose of electroplating is to give a protective coating to the surface beneath. For example, bath taps are chromium plated to prevent corrosion, and at the same time are given a shiny, more attractive finish. The electroplating process is carried out in a cell. This is often known as the ‘plating bath’ and it contains a suitable electrolyte, usually a solution of a metal salt. For silver plating, the electrolyte is a solution of a silver salt. The article to be plated is made the cathode in the cell so that the ions move to it when the current is switched on. [8]
4
a
This increases the conductivity of the water.
b
Chlorine and sodium hydroxide will react together and so the products would not be as wanted.
c
The ions will be mobile in the liquid state.
Supplement 5
6
[Correct ionic charge, correct molecules, correct numbers of electrons] a
Na+ + e– → Na
b
2Br– → Br2 + 2e–
c
Ca2+ + 2e– → Ca
d
Cu2+ + 2e– → Cu
e
2I– → I2 + 2e–
f
4OH– → 2H2O + O2 + 4e–
a
[Correct balancing of ions and electrons]
b
i
2H+(aq) + 2e– → H2(g)
ii
2Cl–(aq) → Cl2 + 2e–
i
1 mole of NaCl is 23 + 35.5 = 58.5 g; so moles of NaCl electrolysed = 234/58.5 = 4 moles From the equation, 2 moles NaCl produces 2 moles NaOH, so 4 moles NaCl produces 4 moles NaOH, or 4 × (23 + 16 + 1) g, i.e. 160 g of NaOH
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Cambridge IGCSE Chemistry Workbook answers
ii
2 moles of NaCl are needed to produce 1 mole Cl2, so 4 moles NaCl produce 2 moles Cl2 , i.e. 2 × (35.5 + 35.5) g = 142 g of Cl2
iii 2 moles of NaCl are needed for 1 mole H2, so 4 moles NaCl produce 2 moles H2, i.e., 2 × (1+1) g H2 = 4 g of H2
Exam-style questions Supplement 1
a
The temperatures required for chemical reduction are too high [1] and therefore energy costs are too high. [1]
b
The mixture of cryolite and aluminium oxide [1] has a much lower melting point, of about 1000°C. [1]
c
i
cathode [1]
ii
Al3+ + 3e– → Al [1]
i
carbon [1]
ii
2O2– → O2 + 2e–
i
The carbon anodes react with oxygen at high temperatures [1], producing carbon dioxide. [1]
ii
C(s) + O2(g) → CO2(g) [1 for reactants, 1 for products]
d e
2
f
It is cheaper to recycle the aluminium than to get it from the ore. [1] It is environmentally more sensible – less mining is required and there is less waste. [1]
a
i
increases [1]
ii
Metallic tin is deposited at the cathode. [1]
i
+2 [1]
ii
The electrolyte is tin(II) sulfate. [1]
i
No [1]
ii
It stays the same because the tin taken out of the solution [1] is replaced by tin from the anode. [1]
i
Sn2+(aq) + 2e– → Sn(s) [1 for reactants, 1 for products, 1 for correct balancing]
ii
Sn(s) → Sn2+(aq) + 2e– [1 for reactants, 1 for products, 1 for correct balancing]
b c
d e
Mild steel is mainly iron, which would corrode when in contact with the food and any solutions in the can. [1]
6 Chemical energetics Core 1
a
Ease of importing oil by sea
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14
Cambridge IGCSE Chemistry Workbook answers
b
i
The crude oil is heated to about 400°C. This vaporises most of the different substances in the crude oil mixture.
ii
naphtha
iii surfacing roads iv diesel oil c
2
i
bitumen
ii
refinery gas
d
gasoline or petrol
e
fractionally distilled
a
A substance that releases energy when it is combusted.
b
i
one from: gasoline/petrol, diesel or other liquid fuel from the fractional distillation of crude oil
ii
e.g. coal
iii e.g. natural gas c
e.g. methane + oxygen → carbon dioxide + water CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) [Correct reactants, correct products, correct balancing]
Supplement 3
a
i
2 × –728 = –1456 kJ
ii
0.25 × –728 = –182 kJ
iii (8/16) × –728 = –364 kJ iv (64/16) × –728 = –2912 kJ
4
b
[Correctly shown axes, correct formulae and balancing, reactants shown higher than products]
a
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Breaking: 2 C—C, + 8 C—H, + 5 O═O: (2 × 347) + (8 × 413) + (5 × 498) = 6488 kJ/mol
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Cambridge IGCSE Chemistry Workbook answers
Forming: 6 C═O, + 8 O—H: (6 × 805) + (8 × 464) = 8542 kJ/mol ∆H = 6488 – 8542 = –2054 kJ/mol b
c
i
[Correctly shown axes, correct formulae and balancing, reactants shown higher than products]
ii
The activation energy, Ea, is the minimum energy that colliding particles must have in order to react.
i
0.5 × –2054 = –1027 kJ
ii
5 × –2054 = –10 270 kJ
iii (11/44) × –2054 = –513.5 kJ 5
a
2H2(g) + O2(g) → 2H2O(l) (or H2(g) + ½O2(g) → H2O(l) ) [Correct reactants, correct products, correct balancing, correct state symbols]
b
endothermic, energy has to be supplied to break chemical bonds
c
If the first equation is given in part a: Breaking: 2 H—H + 1 O═O, (2 × 436) + (1 × 498) = 1370 kJ Forming: 4 O—H, 4 × 464 = 1856 kJ ∆H = 1370 – 1856 = –486 kJ (for 2 moles H2O)
Or, if the second equation is given in part a: Breaking: 2 H—H + 1 O═O, (1 × 436) + (½ × 498) = 685 kJ Forming: 2 O—H, 2 × 464 = 928 kJ ∆H = 685 – 928 = –243 kJ (for 1 mole H2O)
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Cambridge IGCSE Chemistry Workbook answers
6
a
[Correct scale, axis labels, plotting of points]
b i ii
c
–3300 kJ/mol (±150 kJ/mol) [Correct value, correct ‘–' sign] Extrapolated the graph to a relative molecular mass of 88, assuming that the change in enthalpy of the next alcohol which has one extra –CH2 unit would have the same effect on the trend shown.
The amount of energy released per mole gets progressively larger.
Exam-style questions Core 1
a
i
An exothermic reaction is one which transfers thermal energy to the surroundings [1] leading to an increase in the temperature of the surroundings. [1]
ii
A, B, D [1]
iii B [1] iv It has the biggest energy gap between the reactants and products. [1] b
i
An endothermic reaction is one which takes in thermal energy from the surroundings [1] leading to a decrease in the temperature of the surroundings. [1]
ii
E [1]
iii It has the biggest energy gap between the reactants and products. [1]
Supplement 2
a
1.91 g [1]
b
20.8 °C [1]
c
energy transferred = 500 [1] × 4.2 × 20.8 [1] = 43 680 J [1]
d
relative molecular mass of butan-1-ol = 74 [1]
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moles of butan-1-ol burned = 1.91/74 [1] = 0.026 mole [1] e
enthalpy of combustion of butan-1-ol = 43 680/0.026 [1] = 1 680 000 J/mol (1680 kJ/mol) [1 for value, 1 for units]
7 Chemical reactions Core 1
2
a
chemical change
b
physical change
c
physical change
d
chemical change
a
The volume of hydrogen gas collected in a burette/measuring cylinder/gas syringe could be recorded at set time intervals.
b
i
A
ii
Steepest line at the beginning of the curves, or flattens soonest.
iii Yes because all the reactions stop, shown by the flattened lines, to produce the same final volume of hydrogen gas. 3
4
a
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) [Correct reactants, correct products, correct balancing]
b
To prevent the loss of acid spray from the flask, which would have led to increased values for the loss in mass of the flask and its contents.
c
Carbon dioxide gas was being lost from the flask.
d
[Correct scale, correct axis labels, key, points plotted correctly (each graph), each correct best-fit line]
e
The reaction with the smaller chips. The line is steeper at the beginning.
a
It indicates the reaction is reversible.
b
Water could be added.
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c
Hydrated copper(II) sulfate contains water of crystallisation within its structure, anhydrous copper(II) sulfate has no water in its structure.
Supplement 5
a
In all experiments, moles of Mg used = 2/24 = 0.083 mole. Maximum number of moles of H2SO4 used = 0.1 × 40/1000 = 0.004 mole. There is a 1 : 1 mole relationship in the balanced equation, so magnesium is in excess.
b
6
Line
A
B
C
D
E
Experiment
II
III
I
V
IV
c
The reaction shown by line B uses 2 g of powdered magnesium, which has a larger surface area than the magnesium ribbon used in the reaction shown by line C. Hence more collisions occur in the same time, so successful collisions occur more frequently and this gives a faster rate of reaction.
d
The reaction shown by line D is carried out at a higher temperature. So the particles have more energy and move faster, causing successful collisions to occur more frequently and hence giving rise to a faster rate of reaction.
a
2CO(g) + 2NO(g) → 2CO2(g) + N2(g) [correct reactants, correct products, correct balancing]
b
i
carbon monoxide
ii
nitrogen monoxide
i
platinum (alloyed with small amounts of palladium and rhodium)
ii
A catalyst increases the rate of reaction by providing an alternative reaction path with a lower activation energy, so more of the collisions are successful, so increasing the rate of the reaction.
i
C8H18(l) + 12½O2(g) → 8CO2(g) + 9H2O(g) (or 2C8H18 + 25O2 → 16CO2 + 18H2O) [Correct reactants, correct products, correct balancing]
ii
Mass of octane burned = 5000 × 0.70 = 3500 g
c
d
Moles of octane burned = 3500/114 = 30.7 moles Mass of carbon dioxide produced = 30.7 × 44 × 8 = 10 806 g (10.806 kg) iii Volume of carbon dioxide produced = 30.7 × 8 × 24 = 5894.4 dm3 iv 100 g of CO is 100/28 = 3.57 moles; produces 3.57 moles of CO2, this is 3.57 × 44 = 157 g
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Exam-style questions Core 1
a
b
i
Yes. [1]
ii
More oxygen gas is produced, more quickly when more black powder is used. [1] The mass of black powder remains unchanged. [1]
i
At the beginning [1], the line is steepest here. [1]
ii
Maximum concentration of reactants. [1]
iii The reactants are being used up therefore less oxygen gas is produced in a given time. [1] iv 43 cm3 [1] v
48 seconds [1]
Supplement 2
a
To ensure a fair test [1]; different volumes would lead to different concentrations of hydrochloric acid. [1]
b
Solid yellow sulfur is formed in the reaction [1] so it becomes impossible to see the cross. [1]
c Experiment
Rate of reaction / s
1 2
2.3 × 10–2
3
1.5 × 10–2
4
1.0 × 10–2
5
0.6 × 10–2 [1 for each]
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d
[Correct scale, correct axis labels, points plotted correctly, correct best-fit line]
e
i
0.15 mol/dm [1]
ii
45 cm3 [1]
iii 83 s [1] 3
a
Reaction A [1], steepest line [1]
b
i
Reaction A [1]
ii
Reaction C [1]
c
By using powdered zinc or more concentrated acid. [1]
d
Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) [1] Moles of H2 produced = 25/24 000 = 1.04 × 10–3 mole [1] Volume of H2SO4 = 1.04 × 10–3 × 1000/0.05 = 20.8 cm3 [1]
e
Moles of H2 produced = 50/24 000 = 2.08 × 10–3 mole [1] Mass of H2 produced = 2.08 × 10–3 × 2 = 4.16 × 10–3 g [1]
4
a
A reversible reaction, in a closed system reaches an equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction [1]. The concentrations of the reactants and products no longer change. [1]
b
i
Increasing the pressure increases the yield of ammonia [1] at any temperature. [1]
ii
Decreasing the temperature increases the yield of ammonia. [1]
c
As the pressure is increased, the position of equilibrium will move to the right [1] as this produces fewer moles of gas [1], lowering the pressure of the system. [1]
d
The production of ammonia is an exothermic process (negative sign). [1] A decrease in temperature would favour the exothermic reaction. [1] As the yield of ammonia increases with decreasing temperature, the forward reaction must be exothermic. [1]
e
25%
f
At lower temperatures, although more ammonia would be produced [1], the rate would be too slow for the process to be economic. [1]
g
It would be expensive [1] and dangerous. [1]
h
If more nitrogen gas was introduced into the reaction vessel its concentration would be increased [1] and so the equilibrium would move to the right, producing more ammonia. [1]
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5
i
The iron catalyst would not affect the position of the equilibrium, [1] but it would increase the rate of both the forward and reverse reactions equally. [1]
a
i
2SO2(g) + O2(g) ⇌ 2SO3(g) [1 for correctly balanced equation, 1 for reversible arrows]
ii
If the temperature is increased, the position of equilibrium will move to the left. [1] This is because the back reaction is endothermic and this is favoured by higher temperatures. [1] If the pressure is increased, the position of equilibrium will move to the right. [1] This is because there are 3 moles of gas on the left and 2 moles of gas on the right, so moving to the right will form fewer moles of gas, lowering the pressure. [1]
b
By burning sulfur in air, OR from the roasting of sulfide ores. [1]
c
Pressure 200 kPa; [1] Temperature 450oC; [1] Catalyst vanadium(V) oxide. [1]
8 Acids, bases and salts Core 1 Indicator Red litmus
Colour with hydrochloric acid Red
Colour with sodium hydroxide Blue
Blue litmus
Red
Blue
Thymolphthalein
Colourless
Blue
Methyl orange
Red
Yellow
2
ammonium chloride + calcium hydroxide → calcium chloride + ammonia + water [Correct reactants, correct products]
3
a
sodium carbonate + nitric acid → sodium nitrate + water + carbon dioxide Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2 [Correct reactants, correct products, correct balancing]
b
magnesium + hydrochloric acid → magnesium chloride + hydrogen Mg + 2HCl → MgCl2 + H2 [Correct reactants, correct products, correct balancing]
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4 Substances used to make the salt
Salt prepared
Other products
Calcium chloride
Water
Sodium hydroxide/oxide Nitric acid Zinc Lead chloride Barium sulfate
5
a
Potassium chloride
i
Add a small amount of dilute nitric acid followed by a few drops of silver nitrate solution. A white precipitate will be formed.
ii
Add a small amount of dilute nitric acid followed by a few drops of silver nitrate solution. A cream (off-white) precipitate will be formed.
iii Add a small amount of dilute nitric acid followed by a few drops of silver nitrate solution. A yellow precipitate will be formed.
6
b
Add some dilute hydrochloric acid. Fizzing (effervescence) will be observed, caused by the production of carbon dioxide (which would turn limewater cloudy).
c
Add a small amount of dilute hydrochloric acid followed by some barium chloride. A white precipitate will be formed.
d
i
Add some dilute sodium hydroxide solution. A green precipitate will be formed.
ii
Add some dilute sodium hydroxide solution. An orange-brown precipitate will be formed.
a
ammonium
b
soluble, silver
c
soluble
d
sulfates lead
e
insoluble
Supplement 7
a
P is iron(III) chloride, FeCl3
b
Q is iron(III) hydroxide, Fe(OH)3
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8
c
R is iron(II) chloride, FeCl2
d
S is hydrogen, H2
e
T is silver chloride, AgCl
a Substance 0.1 mol/dm3 HCl
pH 1.0
0.1 mol/dm3 NaOH
13.0
0.1 mol/dm3 CH3COOH
2.9
Pure H2O
7.0
0.1 mol/dm3 NH3 solution
11.0
b
Although they have the same concentration, HCl has a lower pH because it is a strong acid and completely dissociates to produce more H+ ions than CH3COOH, which is a weak acid and only partially dissociates to give fewer H+ ions. The higher the concentration of H+ ions in the solution the lower the pH.
c
i
A proton is a hydrogen ion, H+.
ii
An acid is a proton donor.
iii A base is a proton acceptor. 9
a
i
silver nitrate, sodium chloride (or other soluble chloride)
ii
barium chloride or nitrate, sodium sulfate (or other soluble sulfate)
iii calcium nitrate or chloride, sodium or potassium carbonate b
Mix together solutions of lead(II) nitrate and sodium or potassium iodide to give a yellow precipitate of lead(II) iodide. Filter the mixture obtained. Wash with distilled water. Dry the yellow solid. Pb2+(aq) + 2I–(aq) → PbI2(s)
10 a b c d
It produces hydrogen ions (H+(aq)) when added to water. i
an acid that only partially ionises (dissociates) when added to water
ii
an acid that completely ionises (dissociates) when added to water
i
HCl(g) → H+(aq) + Cl–(aq)
ii
CH3COOH(l) ⇌ CH3COO–(aq) + H+(aq)
A strong acid is one that completely ionises (dissociates) when added to water, whereas a weak acid ionises (dissociates) only partially. A concentrated acid contains a high concentration of acid particles, whereas a dilute acid contains a much lower concentration of acid particles and lots of water. [1]
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Exam-style questions Core 1
a
Acids dissolve in water to produce hydrogen ions, which can be written as H+(aq). Bases are oxides or hydroxides of metals. Alkalis are soluble bases. They dissolve in water to produce hydroxide ions, which can be written as OH–(aq). Acids and alkalis react together to produce solutions with a pH of 7; these are called neutralisation reactions. [1 for each]
b
H+(aq) + OH–(aq) → H2O(l) [1 for reactants, 1 for products, 1 for correct balancing]
c
i
2HCl(aq) + K2CO3(s) → 2KCl(aq) + H2O(l) + CO2(g) [1 for reactants, 1 for products, 1 for correct balancing]
ii
To ensure that all the hydrochloric acid had been neutralised. [1]
iii potassium chloride solution [1] iv a saturated solution [1]
Supplement 2
a
Seven (or more) from: •
25 cm3 of the dilute sodium hydroxide solution is placed in a conical flask using a pipette and safety filler. [1]
•
Three or four drops of thymolphthalein indicator are added to the sodium carbonate solution. [1]
•
A burette is filled with the dilute sulfuric acid solution, ensuring that some runs through the valve, and the initial burette reading is taken. [1]
•
The acid is added from the burette into the flask, with swirling, until the colour of the indicator just changes from blue to colourless. [1]
•
The final burette reading is taken and the volume of acid needed to neutralise the sodium hydroxide is found. [1]
•
The process is repeated, this time without an indicator but using exactly the same volumes of dilute sodium hydroxide and dilute sulfuric acid as found in the first experiment. [1]
•
The neutralised solution from the repeat experiment is poured into an evaporating basin and the solution is heated until half of the volume has evaporated. [1]
• b
The solution is then left to cool and the crystals will form slowly. [1]
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) [1 for reactants, 1 for products, 1 for correct balancing]
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9 The Periodic Table Core 1
2
a
chemical change
b
60–100°C 350 – 400 K
c
The melting points increase with increasing atomic number.
d
F and Cl
a
aluminium + chlorine → aluminium chloride 2Al(s) + 3Cl2(g) → 2AlCl3(s)
b
Chlorine is a toxic gas.
c
fluorine
d
any suitable metal, e.g. sodium, potassium, magnesium or gallium
3
The modern Periodic Table has been credited to the work of the Russian chemist Dmitri Mendeleev. After many years of chemists across the world trying to classify the elements in a useful way, he came up with the table that we have been using for nearly 150 years. He arranged the elements in order of increasing atomic weight. Occasionally he had to swap elements around so that they were in the same group as other elements with similar properties, for example swapping tellurium (Te) and iodine (I). The major change that he introduced to his classification was that he left spaces/gaps for elements that had not been discovered at the time. Today, the elements in the modern Periodic Table are arranged in order of increasing atomic number. Going across a period, from left to right, in the modern Periodic Table elements change from being metallic to non-metallic.
4
a
i
19 electrons, 19 protons, 20 neutrons
ii
2,8,8,1
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iii Group I, it has one electron in the outer energy level of the atom iv Y+ b
7
c
i
potassium bromide
ii
potassium + bromine → potassium bromide 2K + Br2 → 2KBr
iii Both chlorine and bromine gain an electron, from the potassium, when they react. Chlorine is more reactive than bromine as it is a smaller atom, which attracts the incoming electron more strongly as its nucleus is closer to its outer energy level. 5
6
a
Mn
b
C
c
Br
d
Na
e
Ne or Kr
f
P
g
Na
h
F
a Group I metals
Transition metals
Density
Low
High
Melting points
Low
High
Colour of solid compounds
White
Coloured
Good catalysts?
No
Yes
b
Higher density than Group I metals, harder/stronger than Group I metals.
c
i
metallic bonding
ii
d
i
It speeds up the chemical reaction and allows lower temperatures/less energy to be used.
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ii
Haber process – iron catalyst, or Contact process – vanadium(V) oxide catalyst
Supplement 7
a b
i
element C: 2,8,3; element D: 2,6
ii
17
i
element A
ii
element B
iii element E iv element C c
i
element C
ii
element D
iii element A
Exam-style questions Core 1
a
potassium bromide + chlorine → potassium chloride + bromine [1] 2KBr(aq) + Cl2(aq) → 2KCl(aq) + Br2(aq) [1 for reactants, 1 for products, 1 for correct balancing]
2
3
b
Chlorine is a smaller atom than bromine. [1] Therefore it attracts the incoming electron from the potassium more strongly as its nucleus is closer to the outer energy level than bromine’s. [1]
c
fluorine [1]
d
No [1], bromine is less reactive than fluorine [1]
a
Na 2,8,1 [1]
b
Potassium. [1] Both potassium and sodium have one electron in their outer energy level, which is lost when they react with the water. [1] In potassium (the bigger atom), the electron in the outer energy level is further from the nucleus and is less tightly held in the atom and so is lost more easily, making it more reactive. [1]
c
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) [1 for reactants, 1 for products, 1 for correct balancing]
d
Francium. [1] It is at the bottom of the group, is the largest atom and loses its outer electron most easily. [1]
a
chlorine: gas, pale yellow-green [1] bromine: liquid, red-brown [1] iodine: solid, green-black [1]
b
7 [1]
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c
gain one electron [1]
d
–1 [1]
e
Fluorine. [1] It is the smallest of all the halogen atoms [1] and attracts the incoming electron most strongly, as its nucleus is closest to the outer energy level into which the electron is coming. [1]
f
i
black [1] solid [1]
ii
Anywhere between –67 and –150 oC [1]
Supplement 4 Formula of compound Oxidation state Name of compound FeCl2
+2 [1]
iron(II) chloride [1]
Fe2O3
+3 [1]
iron(III) oxide [1]
CuO
+2 [1]
copper(II) oxide [1]
CoCl2
+2 [1]
cobalt(II) chloride [1]
10 Metals Core 1
2
Physical property
Metals
Non-metals
Thermal conductivity
High
Low
Electrical conductivity
High
Low
Malleability and ductility
Very
Brittle
Melting and boiling points
High
Low
a
Iron: displacement using carbon. Aluminium: electrolysis. Iron is a moderately reactive metal. Aluminium is a more reactive metal and carbon would not be able to displace aluminium from its ore so a more vigorous method is needed.
b
To produce heat required by reacting with oxygen; to produce carbon monoxide.
c
After decomposing, to form a slag with the sandy impurities from the hematite.
d
[Correct reactants, correct products, for each part] i
calcium carbonate → calcium oxide + carbon dioxide
ii
carbon (coke) + oxygen → carbon dioxide
iii carbon (coke) + carbon dioxide → carbon monoxide Cambridge IGCSE Chemistry Workbook 3rd Edition © Hodder & Stoughton Ltd 2021
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iv iron(III) oxide + carbon monoxide → iron + carbon dioxide calcium oxide + sand (silicon dioxide) → slag (calcium silicate)
v e
[Correct reactants, correct products, correct balancing, for each part] i
CaCO3(s) → CaO(s) + CO2(g)
ii
C(s) + O2(g) → CO2(g)
iii C(s) + CO2(g) → 2CO(g) iv Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) CaO(s) + SiO2(s) → CaSiO3(s)
v 3
f
reducing agent
a
i
carbon
ii
It is reacted with oxygen gas, through a water-cooled lance, to produce oxides that are lost as gases.
i
alloy
ii
They can be harder and stronger than the pure metals and so are more useful.
b
iii hard, resists corrosion c
Stainless steel is too expensive and too dense.
d
4
Object
Properties
Steel
Car body
Easily shaped, not brittle
Mild steel
Axe
Tough
Hard steel
Surgical knife
Tough, sharp-edged, non-reactive
Stainless steel
e
A
a
i
It is a drying agent and removes the water vapour from the air.
ii
Boiling the water removes any oxygen gas from the water because gases are less soluble in hot water.
iii Oil is less dense than water and floats on its surface. It prevents oxygen gas from the air above re-dissolving into the water.
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b Tube
Water
Oxygen
A
✓
✓ ✓
B
C
✓
D
✓
c
d e f
✓
i
tubes B and C
ii
In tube B there is no water and in tube C there is no oxygen. Both water and oxygen are needed for iron to rust.
i
tube D
ii
There is a higher concentration of oxygen gas in tube D than in tube A.
i
A barrier method by painting/greasing/coating with plastic/(galvanising).
ii
Barrier methods prevent rusting by excluding oxygen or water.
i
Zinc is a more reactive metal than iron and it would react rather than the iron.
ii
Galvanising. It acts as a barrier method in that whilst the thin layer is intact it prevents air and water getting to the metal. Because it is a more reactive metal than iron, and loses electrons more easily, the zinc, rather than the iron, reacts with air and water, sacrificially protecting the iron.
Supplement 5
6
a
MgCO3(s) + H2SO4(aq) → MgSO4(aq) + H2O(l) + CO2(g)
b
2Ca(s) + O2(g) → 2CaO(s)
c
Mg(s) + ZnSO4(aq) → MgSO4(aq) + Zn(s)
d
2Mg(s) + O2(g) → 2MgO(s)
e
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
a
2K(s) + 2H2O(l) → 2KOH(aq) + H2(g) [Correct reactants, correct products, correct balancing]
b
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) [Correct reactants, correct products, correct balancing]
c
Mg(s) + CuO(s) → MgO(s) + Cu(s) [Correct reactants, correct products, correct balancing]
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7
8
d
Mg(s) + H2O(g) → MgO(s) + H2(g) [Correct reactants, correct products, correct balancing]
e
Mg(s) + 2H2O(l) → Mg(OH)2(aq) + H2(g) [Correct reactants, correct products, correct balancing]
a
a change in colour of the solution and/or the solid
b
displacement
c
E, F, B, A, D, C
a
copper(II) oxide + zinc → copper + zinc oxide CuO(s) + Zn(s) → Cu(s) + ZnO(s)
b
copper
c
zinc oxide
d
i
zinc
ii
Zinc is gaining oxygen, or the zinc is losing electrons.
iii The copper is being reduced.
Exam-style questions Core 1
a
A, zinc oxide, ZnO; B, magnesium oxide, MgO; C, zinc, Zn; D, zinc chloride, ZnCl2; E, hydrogen gas, H2; F, magnesium, Mg; G, oxygen gas, O2; H, copper, Cu; I, magnesium sulfate, MgSO4 [1 for each]
b
i
zinc oxide + magnesium → magnesium oxide + zinc [1 for reactants, 1 for products]
ii
zinc + hydrochloric acid → zinc chloride + hydrogen [1 for reactants, 1 for products, 1 for correct balancing]
c
Anode: 2O2– → O2(g) + 4e– [1 for reactants, 1 for products, 1 for correct balancing] Cathode: Mg2+ + 2e– → Mg(s) [1 for reactants, 1 for products, 1 for correct balancing]
Supplement 2
a
CuO + Zn → ZnO + Cu [1 for reactants, 1 for products]
b
Cu: +2, 0 [1] Zn: 0, +2 [1]
c
0 [1]
d
Cu2+ + 2e- → Cu [1]
e
reduction [1] it decreases [1]
f
Zn → Zn2+ + 2e- [1]
g
oxidation [1] it increases [1]
h
redox reaction [1]
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3
a
[1 for each] b
In the pure metal all the atoms are of the same size, so when a force is applied, the layers of atoms can easily move over one another. [1] In an alloy the atoms are not all of the same size [1] so when a force is applied the layers of atoms can no longer slide over one another, making the alloy harder and stronger than the pure metal. [1]
11 Chemistry of the environment Core 1
2
3
a
false
b
true
c
false
d
true
e
false
f
false
g
false
h
true
a
Excess fertiliser dissolves and runs off fields into streams and rivers during wet weather.
b
ammonium, NH4+, nitrate, NO3–
c
ammonium nitrate, NH4NO3
d
i
algae and aquatic plants
ii
As algae and aquatic plants die and decay, oxygen is removed from the water. This leaves insufficient oxygen for fish and other organisms to survive.
i
These are gases produced by industries or vehicles and they would not naturally be found in the air.
ii
Any two of these gases:
a
Nitrogen monoxide from the reaction of nitrogen and oxygen, from the air, in the car engine. Nitrogen dioxide from the oxidation of nitrogen monoxide by oxygen from the air. Carbon monoxide from in incomplete combustion of the fuel. Cambridge IGCSE Chemistry Workbook 3rd Edition © Hodder & Stoughton Ltd 2021
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b
i
Fossil fuels, e.g. oil, coal and natural gas
ii
64 kg
iii sulfurous acid, H2SO3, sulfuric acid, H2SO4 iv Extensive damage to forests, increased corrosion of exposed metals; and damage to buildings and statues made from limestone and marble 4
a
v
flue gas desulfurisation units or FGD units
i
The water is passed through screens.
ii
To filter out floating debris.
b
To kill any remaining bacteria in the water.
c
hydrochloric acid, HCl
d
To neutralise the acidic solution.
e
Removes colour and smells.
f
fluoride, F–
Supplement 5
a
i
100 × 14/(23 + 14 + 48) = 16.47%
ii
100 × (14 × 3)/[3(14 + 4) + (31+64)] = 28.19%
iii 100 × (14 × 2)/[2(14 + 4) + (32+64)] = 21.21%
6
b
potassium (K) and phosphorus (P)
c
Ca3(PO4)2
a
i
1 mole CH4 is 12 + (4 × 1) = 16 g, and produces 1 mole CO2, 12 + (2 × 16) = 44 g 32 g of CH4 = 2 moles; produces 2 moles, i.e., 88 g of CO2
ii
88 g of CO2 = 2 moles of CO2; it occupies 2 × 24 dm3 = 48 dm3
b
C8H18 contains 8 times as much carbon as is found in methane.
Exam-style questions Core 1
a
i
a gas that contributes to the greenhouse effect [1] by absorbing infrared radiation [1], leading to atmospheric warming
ii
a general warming [1] across the surface of the Earth caused by the greenhouse effect [1]
b
climate changes [1], melting of the ice caps [1], flooding caused by rise in sea levels [1]
c
Methane [1], CH4 [1]
d
Plant more trees [1], halt deforestation [1], use less fossil fuel for the production of electricity [1], use fewer/no petrol or diesel vehicles by using electric ones [1], increase the use of alternative energy sources [1], build houses that are more energy-efficient, [1] or other suitable ways.
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Cambridge IGCSE Chemistry Workbook answers
Supplement 2
a
i
96 dm3 [1]
ii
2NO(g) + O2(g) → 2NO2(g)
iii catalytic converter [1] b
i
nitric acid [1], HNO3 [1]
ii
nitrogen(IV) oxide + water → nitric acid + nitrogen monoxide [1] 3NO2(g) + H2O(g) → 2HNO3(aq) + NO(g) [1 for reactants, 1 for products, 1 for correct balancing]
12 Organic Chemistry 1 Core 1
2
a
The process of breaking large molecules into smaller, more useful molecules.
b
A substance that alters the rate of a chemical reaction without itself being chemically changed at the end of the reaction.
c
The chemical breakdown of a substance under the influence of heat.
d
A family of saturated hydrocarbons with the general formula CnH2n+2.
e
Molecules that possess only single covalent bonds.
f
Molecules that contain one or more double covalent bonds.
g
An atom or group of atoms that determine the chemical properties of a homologous series.
h
Compounds that contain carbon and hydrogen only.
a
When small molecules such as ethene join together to form long chains of atoms, called polymers, the process is called polymerisation. These small molecules that can join together in this way are called monomers. The polymer formed with ethene is an addition polymer.
b
Single C—C bonds and shows at least 3 molecules.
3
c
The colour changes from orange to colourless.
d
Some plastics chemically break down during the recycling process. Hence they cannot be recycled.
a
i
addition polymerisation
ii
poly(ethene)
iii CnH2n
Cambridge IGCSE Chemistry Workbook 3rd Edition © Hodder & Stoughton Ltd 2021
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Cambridge IGCSE Chemistry Workbook answers
b
i
The plastic does not react with substances in the environment and is not decomposed by bacteria, so it will not rot away. The bags will therefore stay unchanged over many, many years. They create a polluted environment and are a danger to animals.
ii
In recycling, new bags are made from the existing plastic, saving the costs of energy and raw materials.
iii Not all plastics just melt, some break down when heated. iv Landfill sites are getting full. Plastics are accumulating in the oceans. Plastics form toxic gases when burned. 4
a
i
Homologous series: a homologous series is a family of similar compounds, with similar chemical properties, due to the presence of the same functional group, with the same general formula, differing by CH2. Hydrocarbons: molecules that contain hydrogen and carbon atoms only.
ii b
Combustion and substitution with chlorine.
i Alkane
Formula
Methane
CH4
Structure
Ethane
Propane
C3H8
Butane
C4H10
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36